Suppose we have a Sturm-Liouville differential equation $$y''-2xy'+\lambda y=0.$$ The equation has a polynomial solution in the case $\lambda =4$. So if we write the equation in self-adjoint form, we get: $$\frac{d}{dx}\bigg(e^{-x^2}\frac{dy}{dx}\bigg)+4e^{-x^2}y=0. \qquad(2)$$
This implies that the polynomial solution and $\lambda=4$ are the eigenfunction and eigenvalue of the corresponding eigenvalue problem, $$\hat{L}[y]=4\omega(x)y. \qquad(3)$$ My question is about the implication, I do not understand how from equation $(2)$ we can imply that equation $(2)$ is actually the eigenfunction of equation $(3)$.
I would appreciate any help or suggestion. Thank you.
If you consider the operator $L$ as
$$L=\frac{1}{w}\frac{d}{dx}[w\frac{d}{dx}]$$
where $w(x)=e^{-x^2}$ is a weight function then you can see that
$$L[y]=-\lambda y$$
I think what made this hard for you to see it as the usual eigenvalue problem in matrices $Ax=\lambda x$, was the presence of the weight function $w$ which you didn't include it in the operator $L$.